NCERT Solutions for Class 9 Maths Chapter 11 Construction

Ans : The below given steps will be followed to construct an angle of 900.
(i) Take the given ray PQ. Draw an arc of some radius taking point P as its centre, which intersects PQ at R.
(ii) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn att at S.
(iii) Taking S as centre and with the same radius as before, draw an arc intersecting the arc at T (see figure).
(iv) Taking S and T as centre, draw an arc of same radius to intersect each other at U.
(v) Join PLJ, which is the required ray making 900 with the given ray PQ.

Justification of Construction:
We can justify the construction, if we can prove \(\angle \mathrm{UPQ}=90^{\circ}\)
For this, join PS and PT.

\(\begin{array}{l}{\text { We have, } \angle \mathrm{SPQ}=\angle \mathrm{TPS}=60^{\circ} . \text { In (iii) and (iv) steps of this construction, PU was }} \\ {\text { drawn as the bisector of } \angle \mathrm{TPS} \text { . }}\end{array}\)
\(\begin{array}{l}{\therefore \angle \mathrm{UPS}=\frac{1}{2} \angle \operatorname{TPS}=\frac{1}{2} \times 60^{\circ}=30^{\circ}} \\ {\text { Also, } \angle \mathrm{UPQ}=\angle \mathrm{SPQ}+\angle \mathrm{UPS}} \\ {=60^{\circ}+30^{\circ}} \\ {=90^{\circ}}\end{array}\) 

Ans : The below given steps will be followed to construct an angle of 450.
(i) Take the given ray PQ. Draw an arc of some radius taking point P as its centre, which intersects PQ at R.
(ii) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S.
(iii) Taking S as centre and With the same radius as before, draw an arc intersecting the arc at T (see figure).
(iv) Taking S and T as centre, draw an arc of same radius to intersect each other at U.
(v) join PU. Let it intersect the arc at point V.
(vi) From R and V, draw arcs with radius more than 1/2 RV to intersect each other at W. Join PW.
PW is the required ray making 450 with PQ.

Justification of Construction:
We can justify the construction, if we can prove \(\angle \mathrm{WPQ}=45^{\circ}\)
For this, join PS and PT.

\(\begin{array}{l}{\text { We have, } \angle \mathrm{SPQ}=\angle \mathrm{TPS}=60^{\circ} . \text { In (iii) and (iv) steps of this construction, PU was }} \\ {\text { drawn as the bisector of } \angle \mathrm{TPS} \text { . }}\end{array}\)
\(\begin{array}{l}{\therefore \angle U P S=\frac{1}{2} \angle \operatorname{TPS}=\frac{60^{\circ}}{2}=30^{\circ}} \\ {\text { Also, } \angle U P Q=\angle S P Q+\angle U P S} \\ {=60^{\circ}+30^{\circ}} \\ {=90^{\circ}}\end{array}\)
In step (vi) of this construction, PW was constructed as the bisector of \(\angle\)UPQ.
\(\therefore \angle \mathrm{WPQ}=\frac{1}{2} \angle \mathrm{UPQ}\) \(=\frac{90^{\circ}}{2}=45^{\circ}\) 

Ans : (i)300
The below given steps will be followed to construct an angle of 300.
Step I: Draw the given ray PQ. Taking P as centre and with some radius, draw an arc of a circle which intersects PQ at R.
Step II: Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point S.
Step III: Taking R and S as centre and with radius more than 2 RS, draw arcs to intersect each other at T. Join PT which is the required ray making 300 with the given ray PQ.



(ii) \(22 \frac{1}{2} ^ \circ\)
The below given steps will be followed to construct an angle of \(22 \frac{1}{2} ^\circ\).
(1) Take the given ray PQ. Draw an arc of some radius, taking point p as its centre, which intersects PQ at R.
(2) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S.
(3) Taking S as centre and with the same radius as before, draw an arc intersecting the arc at T (see figure).
(4) Taking S and T as centre, draw an arc of same radius to intersect each other at U.
(5) Join Let it intersect the arc at point V.
(6) From R and V, draw arcs with radius more than \(\frac{1}{2}\)RV to intersect each other at W. Join PW.
(7) Let it intersect the arc at X. Taking X and R as centre and radius more than \(\frac{1}{2}\)RX, draw arcs to intersect each other at Y.


Joint PY which is the required ray making \(22 \frac{1}{2} ^ \circ\) with the given ray PQ.



(iii) \(15^{\circ}\)
The below given steps will be followed to construct an angle of \(15^{\circ}\).
Step I: Draw the given ray PQ. Taking P as centre and with some radius, draw an arc of a circle which intersects PQ at R.
Step II: Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point S.
Step Ill: Taking R and S as centre and with radius more than 1/2 RS, draw arcs to intersect each other at T. Join PT.
Step IV: Let it intersect the arc at U. Taking IJ and R as centre and with radius more than 1/2 RU, draw an arc to intersect each other at V. Join PV which is the required ray making \(15^{\circ}\) with the given ray PQ.
 

Ans : (i) \(75^{\circ}\)
The below given steps will be followed to construct an angle of \(75^{\circ}\).
(1) Take the given ray PQ. Draw an arc of some radius taking point p as its centre, which intersects PQ at R.
(2) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S.
(3) Taking S as centre and with the same radius as before, draw an arc intersecting the arc at T (see figure).
(4) Taking S and T as centre, draw an arc of same radius to intersect each other at U.
(5) Join PLI. Let it intersect the arc at V. Taking S and V as centre, draw arcs with radius more than 1/2 SV. Let those intersect each other at W. Join PW which is the required ray making \(75^{\circ}\) with the given ray PQ.

The angle so formed can be measured with the help of a protractor. It comes to be \(75^{\circ}\).


(ii) \(105^{\circ}\)
The below given steps will be followed to construct an angle of  \(105^{\circ}\)
(1) Take the given ray PQ. Draw an arc of some radius taking point p as its centre, which intersects PQ at R.
(2) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S.
(3) Taking S as centre and with the same radius as before, draw an arc intersecting the arc at T (see figure).
(4) Taking S and T as centre, draw an arc of same radius to intersect each other at U.
(5) Join PLI. Let it intersect the arc at V. Taking T and V as centre, draw arcs with radius more than 1/2 TV. Let these arcs intersect each other at W. Join PW which is the required ray making \(105^{\circ}\) with the given ray PQ.

The angle so formed can be measured with the help of a protractor. It comes to be \(105^{\circ}\).


(iii) \(135^{\circ}\)
The below given steps will be followed to construct an angle of \(135^{\circ}\)
(1) Take the given ray PQ. Extend PQ on the opposite side of Q. Draw a semi-circle of some radius taking point P as its centre, which intersects PQ at R and W.
(2) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S.
(3) Taking S as centre and with the same radius as before, draw an arc intersecting the arc at T (see figure).
(4) Taking S and T as centre, draw an arc of same radius to intersect each other at U.
(5) Join PU. Let it intersect the arc at V. Taking V and W as centre and with radius more than 1/2 VW, draw arcs to intersect each other at X. Join PX, which is the required ray making \(135^{\circ}\) with the given line PQ.

The angle so formed can be measured with the help of a protractor. It comes to be \(135^{\circ}\). 

Ans : Let us draw an equilateral triangle of side 5 cm. We know that all sides of an equilateral triangle are equal. Therefore, all sides of the equilateral triangle will be 5 cm. We also know that each angle Of an equilateral triangle is \(60^{\circ}\).
The below given steps Will be followed to draw an equilateral triangle Of 5 cm side.
Step I: Draw a line segment AB of 5 cm length. Draw an arc of some radius, while taking A as its centre. Let it intersect AB at P.
Step II: Taking P as centre, draw an arc to intersect the previous arc at E. join AE.
Step Ill: Taking A as centre, draw an arc of 5 cm radius, which intersects extended line segment AE at C. Join AC and BC. \(\Delta\)ABC is the required equilateral triangle of side 5 cm.

Justification of Construction:
We can justify the construction by showing ABC as an equilateral triangle i.e., AB =
\(\begin{array}{l}{\mathrm{BC}=\mathrm{AC}=5 \mathrm{cm} \text { and } \angle \mathrm{A}=\angle \mathrm{B}=\angle \mathrm{C}=60^{\circ} .} \\ {\text { In } \triangle \mathrm{ABC}, \text { we have } \mathrm{AC}=\mathrm{AB}=5 \mathrm{cm} \text { and } \angle \mathrm{A}=60^{\circ} .} \\ {\text { since } \mathrm{AC}=\mathrm{AB}_{\text { }}} \\ {\angle \mathrm{B}=\angle \mathrm{C} \text { (Angles opposite to equal sides of a triangle) }}\end{array}\)
\(\begin{array}{l}{\text { In } \triangle A B C,} \\ {\angle A+\angle B+\angle C=180^{\circ}(\text { Angle sum property of a triangle) }}\end{array}\)
\(\begin{array}{l}{\angle 60^{\circ}+\angle C+\angle C=180^{\circ}} \\ {\angle 60^{\circ}+2 \angle C=180^{\circ}} \\ {\angle 2 \angle C=180^{\circ}-60^{\circ}=120^{\circ}} \\ {\angle C=60^{\circ}} \\ {\angle B=\angle C=60^{\circ}}\end{array}\)
\(\begin{array}{l}{\text { We have, } \angle A=\angle B=\angle C=60^{\circ} \ldots(1)} \\ {\angle A=\angle B \text { and } \angle A=\angle C}\end{array}\)
\(\begin{array}{l}{\angle \mathrm{BC}=\mathrm{AC} \text { and } \mathrm{BC}=\mathrm{AB}(\text { Sides opposite to equal angles of a triangle) }} \\ {\angle \mathrm{AB}=\mathrm{BC}=\mathrm{AC}=5 \mathrm{cm} \ldots(2)}\end{array}\)
from equations (1) and (2), \(/Delta\)ABC is an equilateral triangle.